Interval of Convergence for Power Series

[ineedhelpnow]

Hi hi,
So I worked on this problem and I know I probably made a mistake somewhere towards the end so I was hoping one of you would catch it for me. Thank you!

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[MarkFL]

According to W|A, the series indeed converges for:

\(\displaystyle |-x-1|<3\)

which gives:

\(\displaystyle -4<x<2\)

just as you found. (Yes)

By the way, I am going to move this thread to our Calculus forum as its a better fit I think. (Thinking)

 

[ineedhelpnow]

Thank you ^^ I wasn't sure if it would be -3<-x-1<3 or -3<x+1<3

 

[MarkFL]

Thank you ^^ I wasn't sure if it would be -3<-x-1<3 or -3<x+1<3

Note that:

\(\displaystyle |-x-1|<3\)

is the same as:

\(\displaystyle |x+1|<3\)

which can be written as:

\(\displaystyle \sqrt{(x-(-1))^2}<3\)

and this tells us (because it is a 1-dimensional distance formula) that we want all real $x$ whose distance (on the number line) from $-1$ is less than $3$, and so we have:

\(\displaystyle -1-3<x<-1+3\)

or

\(\displaystyle -4<x<2\)

You should also observe that either inequality you posted will lead to the correct answer:

i) \(\displaystyle -3<-x-1<3\)

Add though by 1:

\(\displaystyle -2<-x<4\)

Multiply through by -1:

\(\displaystyle 2>x>-4\)

And this is equivalent to:

\(\displaystyle -4<x<2\)

ii) \(\displaystyle -3<x+1<3\)

Subtract through by 1:

\(\displaystyle -4<x<2\)

 

[ineedhelpnow]

I noticed they both come to the same solution but I wasn't sure if there was a "right" way to have it done. Thanks a ton for the thorough explanation!